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Forums - General - 0.9999.... = 1.0

 

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Yes 34 58.62%
 
No 20 34.48%
 
not sure 1 1.72%
 
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SuperMarioWorld on 30 April 2013

your logic can't prove -0.99999... = -1



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Soleron on 30 April 2013
SuperMarioWorld said:
Jay520 said:
SuperMarioWorld said:

now do it for 1/8.

it will only work if x is a positive integer or x is 1/9. there is your flaw. therefore you just prooved 0.999999 doesn't equal 1.

yeah, no 

x = (1/8) given
10x = 10/8 multiply by 10
9x = 9/8 subtract x
x = 1/8 divide by 9
1/8 = 1/8 substitution

try again

oops i'm sorry i meant -1/8. your proof only applies to postive numbers therefore has many holes in your logic. However i was mistaken it would only work with positive integers. cheers for that

x = -1/8
10x = -10/8
9x = -9/8
x=-1/8
-1/8=-1/8

In fact I can prove it works for all numbers

let y = x

(10y - y)/9 = y

y = y

x = x



Jay520 on 30 April 2013
SuperMarioWorld said:

oops i'm sorry i meant -1/8. your proof only applies to postive numbers therefore has many holes in your logic. However i was mistaken it would only work with positive integers. cheers for that

 

yeah, no 

x = -1/8 given
10x = -10/8 multiply by 10
9x = -9/8 subtract x
x = -1/8 divide by 9
-1/8 = -1/8 substitution

try again, and please: check before you do so.

 

BTW, 1/8 is not a positive integer.



drkohler on 30 April 2013
fordy said:


Rule 1: x/x = 1
Rule 2: 0/x = 0

In which case, which rule applies to 0/0?

No rule at all. 0/0 is an indeterminate number. Or are you talking about limes x->0?



Zkuq on 30 April 2013
SuperMarioWorld said:
Zkuq said:
SuperMarioWorld said:
Zkuq said:
SuperMarioWorld said:

the step 'subtract x' is a trick. you are treating x as a constant not a variable. so when you say subtract x it means subtract 0.99999 therefore your proof doesn't really hold up mathematically. its a trick

x = 0.999... and therefore 'subtract x' is the same thing as 'subtract 0.999...'. You can operate on both sides of an equation; the form you operate on each side is up to you to decide as long as both forms are equal. Besides, your point about treating x as a constant instead of a variable doesn't make much sense. x is x and it has a known value; what's the problem?

Oh I get it. You're just trying to mess with us.


i'm not messing with you. the problem with his proof is he is treating x as a variable. but it is a constant as he has stated in the first step. you can't just suddenly decide x is a variable in the middle of a proof when he has stated it's a constant at the start of a proof. In a proof you must define x to a degree for example x is any positive integer. he defined x as 0.99999. nothing else. his proof is completely wrong.

http://en.wikipedia.org/wiki/User:ConMan/Proof_that_0.999..._does_not_equal_1 

If you think the proof is wrong, please do point out the step that's wrong and why. All you have is an ambiguous statement about what's wrong but if there's something wrong, you should point it out exactly because it's not obvious. What does a variable even mean to you? And what about the other proofs? Also, have you studied mathematics? Besides high school level, that is.


I did point out the wrong step in my first post mate. 

No you didn't. As far as I'm concerned, all you did point out was your lack of basic understanding of manipulating equations. That, or my English is rusty. What do you even mean when you say he's treating it as a constant instead of a variable? Then let's assume that's a mistake: What does it affect and where?



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ebw on 30 April 2013
dsgrue3 said:
x = 2
x^2 = 4
x = +/- 2

Fuzzy math is fuzzy. This is why it doesn't sit well with me.

I understand it through Pezus' example with fractions, but this isn't an equation. It's just an assertion.

It isn't a proof at all.

Surely you admit -2 != 2?

You are deeply misguided about what a proof is.  Your calculation begins with the premise that x=2 and deduces that x = ±2, which I completely agree with.  You seem to misunderstand what the ± notation means.  The last sentence reads "x must be either 2 or -2", which of course it is, since we happen to know it is 2.  There is nothing in the sentence "x = ±2" that says that x could be -2, only that x cannot take any other value.  Likewise, Pezus's example seeks to show that if x has any value whatsoever, it must be 1.  Reading this as "-2 = 2" is utter nonsense.



fordy on 30 April 2013
drkohler said:
fordy said:


Rule 1: x/x = 1
Rule 2: 0/x = 0

In which case, which rule applies to 0/0?

No rule at all. 0/0 is an indeterminate number. Or are you talking about limes x->0?

Exactly my point. Special rules apply to 0 and Infinite, that set them apart from other numbers.

lim y -> 0 : x/y -> Infinite. That's a another rule that this could apply to. What would 0/0 approach as the denominator approaches 0?



luiaots on 30 April 2013

I feel sorry for people who think they know math when they really really really don't. I bet most people who claim that .999... doesn't equal 1 haven't taken any advanced math classes.



I am a nintendo fan, not a nintendo fanboy

drkohler on 30 April 2013
fordy said:
drkohler said:
fordy said:


Rule 1: x/x = 1
Rule 2: 0/x = 0

In which case, which rule applies to 0/0?

No rule at all. 0/0 is an indeterminate number. Or are you talking about limes x->0?

Exactly my point. Special rules apply to 0 and Infinite, that set them apart from other numbers.

lim y -> 0 : x/y -> Infinite. That's a another rule that this could apply to. What would 0/0 approach as the denominator approaches 0?

I really would have to dig out my old math lecture notes to be sure. I'd guess Hopital rule could apply to Rule 2 so lim x->0  0/x = 0



Zkuq on 01 May 2013
fordy said:
drkohler said:
fordy said:


Rule 1: x/x = 1
Rule 2: 0/x = 0

In which case, which rule applies to 0/0?

No rule at all. 0/0 is an indeterminate number. Or are you talking about limes x->0?

Exactly my point. Special rules apply to 0 and Infinite, that set them apart from other numbers.

lim y -> 0 : x/y -> Infinite. That's a another rule that this could apply to. What would 0/0 approach as the denominator approaches 0?

0/0 doesn't approach anything, it's not defined (not even by your rules because they contradict each other). As for your limit, it doesn't really make much sense for your point, I believe. If x=0, your limit is 0 by rule 2 of yours, and if x!=0, your limit is +/- infinite. I believe the limit you are trying to find is lim (x,y)->(0,0) x/y? Anyway, no solution for that limit because, well, there's no limit there.